2.3 characterizations of invertible matrices

If the nxn matrices E and F have the property that EF = I, then E and F commute. Explain why.

According to the IMT, E and F must be invertible and inverses. So FE = I and I = EF. Thus, E and F commute

If the columns of A span Rn, then the columns are linearly independent

True, the IMT states that if the columns of A span Rn, then matrix A is invertible. Therefore, the columns are linearly independent

If the equation Ax = 0 has only the trivial solution, then A is row equivalent to the nxn identity matrix

True, by the invertible matrix theorem, if the equation Ax = 0 has only the trivial solution, then the matrix is invertible. Thus A must also be equivalent to the nxn identity matrix

If there is an nxn matrix D such that AD = I, then there is also an nxn matrix C such that CA = I

True, by the invertible matrix theorem, if there is an nxn matrix D such that AD = I, then it must be true that there is also an nxn matrix C such that CA = I

If the columns of a 7x7 matrix D are linearly independent, what can you say about the solutions of Dx = b?

Equation Dx = b has a solution for each b in R7. According to the IMT, a matrix is invertible if the columns of the matrix form a linearly independent set, this would mean that the equation Dx = b has at least one solution for each b in Rn

If A is an nxn matrix then the equation Ax=b has least one solution for each b in Rn

False, by the IMT, Ax = b has at least one solution for each b in Rn only if a matrix is invertible

If the linear transformation x->Ax maps Rn into Rn then A has n pivot positions

False, the linear transformation x -> Ax will always map Rn into Rn for any nxn matrix. According to the IMT, A has n pivot positions only if x -> Ax maps Rn onto Rn

Explain why the columns of A^2 span Rn whenever the columns of an nxn matrix are linearly independent

If the columns of A are linearly independent and A is square, then A is invertible, by the IMT. Thus A^2, which is the product of invertible matrices, is also invertible. So, the columns of A^2 span Rn

If A is invertible, then the columns of A^-1 are linearly independent

It is a known theorem that if A is invertible then A^-1 must also be invertible. According to the IMT, if a matrix is invertible its columns form a linearly independent set. Therefore, the columns of A^-1 are linearly independent

The Invertible Matrix Theorem

Let A be a square n x n matrix. Then the following statements are equivalent. That is, for a given A, the statements are either all true or all false. a) A is an invertible matrix b) A is row equivalent to the n x n identity matrix c) A has n pivot positions d) The equation Ax = 0 has only the trivial solution e) The columns of A form a linearly independent set f) The linear transformation x --> Ax is one-to-one g) The equation Ax = b has at least one solution for each B in R^n // The equation Ax = b has a unique solution for each b in R^n h) The columns of A span R^n i) The linear transformation x --> Ax maps R^n onto R^n j) There is an n x n matrix C such that CA = I k) There is an n x n matrix D such that AD = I l) A^T is an invertible matrix

Let A and B be nxn matrices. Show that if AB is invertible so is B

Let W be the inverse of AB. Then WAB = I and (WA)B = I. Therefore, matrix B is invertible by part (j) of the IMT

Can a square matrix with two identical rows be invertible?

No, if a matrix has two identical columns then its columns are linearly dependent. According to the IMT this makes the matrix not invertible

Is it possible for a 5x5 matrix to be invertible when its columns do not span R5?

No. According to the IMT an nxn matrix cannot be invertible when its columns do not span Rn

If C is 6x6 and the equation Cx = v is consistent for every v in R6, is it possible that for some v, the equation Cx = v has more than one solution?

No. Since Cx = v is consistent for every v in R6, according to the IMT that makes the 6x6 invertible. Since it is invertible, Cx = v has a unique solution

If the given equation Gx = y has more than one solution for some y in Rn, can the columns of G span Rn? Why or why not? Assume G is nxn.

The columns of G cannot span Rn. According to the IMT, if Gx = y has more than one solution for some y in Rn, that makes the matrix G non invertible.

If an nxn matrix K cannot be row reduced to I_n, what can you say about the columns of K?

The columns of K are linearly dependent and the columns do not span Rn. According to the invertible matrix theorem, if a matrix cannot be row reduced to I_n that matrix is non invertible

Suppose H is an nxn matrix. If the equation Hx = c is inconsistent for some c in Rn, what can you say about the equation Hx = 0?

The statement that Hx = c is inconsistent for some c is equivalent to the statement that Hx = c has no solution for some c. From this, all of the statements in the IMT are false, including the statement that Hx = 0 has only the trivial solution. Thus, Hx = 0 has a nontrivial solution.

If L is nxn and the equation Lx = 0 has the trivial solution, do the columns of L span Rn?

This fact gives no info about the columns of L. The equation Lx = 0 always has the trivial solution

If there is a b in Rn such that the equation Ax = b is inconsistent, then the transformation x -> Ax is not one to one

True, according to the IMT, if there is a b in Rn such that the equation Ax = b is inconsistent, then equation Ax = b does not have at least one solution for each b in Rn and this makes A not invertible

If A^T is not invertible, then A is not invertible

True, by the IMT, if A^T is not invertible, then all statements in the theorem are false, including A is invertible. Therefore, A is not invertible

If the equation Ax = b has at least one solution for each b in Rn, then the solution is unique for each b

True, by the IMT, if Ax = b has at least one solution for each b in Rn, then matrix A is invertible. If A is invertible, then according to the IMT the solution is unique for each b

If the columns of A are linearly independent, then the columns of A span Rn

True, by the IMT, if the columns of A are linearly independent, then the columns of A must span Rn

If the equation Ax = 0 has a nontrivial solution, then A has fewer than n pivot positions.

True, by the IMT, if the equation AX = 0 has a nontrivial solution, then matrix A is not invertible. Therefore, A has fewer than n pivot positions

Suppose T and U are linear transformations from Rn to Rn such that T(Ux) = x for all x in Rn. Is it true that U(Tx) = x for all x in Rn?

Yes, it is true AB is the standard matrix of the mapping x -> T(U(x)) due to how matrix multiplication is defined. By hypothesis, this mapping is the identity mapping, so AB = I. Since both A and B are square and AB = I, the IMT states that both A and B invertible, and B = A^-1. Thus, BA = I. This means that the mapping x -> U(T(x)) is the identity mapping. Therefore, U(T(x)) = x for all x in Rn.